Browse Questions

Find the values of the expression $\tan \bigg( \sin^{-1} \frac{3}{5} + \cot^{-1} \frac{3}{2} \bigg)$

$\begin{array}{1 1} \frac{17}{6} \\ \frac{7}{6} \\ \frac{3}{2} \\ \frac{-3}{2} \end{array}$

Toolbox:
• $sin^{-1}x=tan^{-1}\large \bigg(\frac{x}{\sqrt{1-x^2}}\bigg)$
• $cot^{-1}x=tan^{-1}\frac{1}{x}$
• $tan^{-1}x+tan^{-1}y=tan^{-1}\frac{x+y}{1-xy}\:\:xy<1$
Given $tan \bigg( sin^{-1} \frac{3}{5} + cot^{-1} \frac{3}{2} \bigg)$
We know that $sin^{-1}x=tan^{-1}\bigg( \large \frac{x}{\sqrt{1-x^2}}\bigg)$
By substituting $\:x=\frac{3}{5},\:\large \frac{x}{\sqrt{1-x^2}}=\frac{\frac{3}{5}}{\sqrt{1-\frac{9}{25}}}=\frac{3}{4}$
$\Rightarrow$ $sin^{-1}\frac{3}{5}=tan^{-1}\frac{3}{4}$
We know that $cot^{-1}x=tan^{-1}\frac{1}{x}$
By Substituting $x=\frac{3}{2}$ we get $\;cot^{-1}\frac{3}{2}=tan^{-1}\frac{2}{3}$
The given expression becomes $tan \bigg[ tan^{-1}\frac{3}{4}+tan^{-1}\frac{2}{3}\bigg]$
We know that $tan^{-1}x+tan^{-1}y=tan^{-1}\frac{x+y}{1-xy}\:\:xy<1$
By taking $x=$$\frac{3}{4}, \:y=\frac{2}{3}$, let us evaluate $tan^{-1}\frac{3}{4}+tan^{-1}\frac{2}{3}$
$x+y = \frac{3}{4} + \frac{2}{3} = \frac{3 \times 3 + 2 \times 4}{4 \times 3} = \frac{17}{12}$
$1-xy = 1 - \frac{3}{4} \times \frac{2}{3}\ = 1 - \frac{1}{2} = \frac{1}{2}$
Therefore, $tan^{-1}\frac{3}{4}+tan^{-1}\frac{2}{3} = tan^{-1} \large\frac{ \frac{17}{12}}{\frac{1}{2}}$$= tan^{-1} \frac{17}{6}$
Therefore, $tan\; ( tan^{-1} \frac{17}{6}) = \large \frac{17}{6}$
edited Mar 14, 2013